EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 8220101443908
Author: Harris
Publisher: MAC HIGHER
Question
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Chapter 15, Problem 15.1P

(a)

Interpretation Introduction

Interpretation:

Half-reaction for the Cu electrode has to be written.

Concept Introduction: Cell potential is evaluated by expression as follows:

  Ecell°=Eoxidation°+Ereduction°

Expression to compute Ecell as per the Nernst equation is written as follows:

  Ecell=Ecell°0.0592nlog[P][R]

Here,

  • Ecell denotes overall cell potential.
  • Ecell° denotes standard cell potential.
  • n is the moles of electrons transferred in each reaction.
  • [P] denotes concentration of products.
  • [R] denotes concentration of reactant.

(a)

Expert Solution
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Explanation of Solution

Since Cu electrode is attached to postive terminal of potentiometer it serves as cathode and therefore reduction of Cu occurs at this electrode.

Reduction half reaction for the Cu electrode is therefore written as follows:

  Cu2++2eCu(s)

(b)

Interpretation Introduction

Interpretation:

Nernst equation corresponding to Cu electrode has to be written.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
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Explanation of Solution

Expression to compute Ecell as per the Nernst equation is written as follows:

  Ecell=Ecell°0.0592nlog[P][R]        (1)

Since Cu(s) is formed as product, value of [Cu] is taken as unity.

Substitute 2 for n, [Cu2+] for [R] and 1 for [P] in equation (1).

  Ecell=Ecell°0.05922log1[Cu2+]        (2)

(c)

Interpretation Introduction

Interpretation:

Cell voltage has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Substitute 0.339 V for Ecell° and 0.10 M for [Cu2+] in equation (2).

  Ecell=0.339 V0.05922log(10.10 M)=0.309 V

Cell potential is evaluated by expression as follows:

  Ecell°=Eoxidation°+Ereduction°        (3)

Substitute 0.339 V for Ereduction° and 0.197 V for Eoxidation° in equation (3).

  Ecell°=0.197 V+0.339 V=0.112 V

Thus, cell potential is 0.112 V.

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